Question: Find $\dfrac{d}{dx}\left[\sec^2(4x)\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $8\sec(4x)$ (Choice B) B $8\sec(4x)\tan(4x)$ (Choice C) C $8\sec^2(4x)\tan(4x)$ (Choice D) D $8\sec^2(4x)$
Answer: $\sec^2(4x)$ is a composition of three functions! Let... $u(x)=x^2$ $v(x)=\sec(x)$ $w(x)=4x$... then $\sec^2(4x)=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $\dfrac{d}{dx}\left[\sec^2(4x)\right]$, we will need to use the chain rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=2x$ $v'(x)=\sec(x)\tan(x)$ $w'(x)=4$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={2\sec(4x)}\cdot{ \sec(4x)\tan(4x)}\cdot{4} \\\\ &=8\sec^2(4x)\tan(4x) \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[\sec^2(4x)\right]=8\sec^2(4x)\tan(4x)$.